Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 4}{x + 1} = \dfrac{4x + 1}{x + 1}$
Multiply both sides by $x + 1$ $ \dfrac{x^2 - 4}{x + 1} (x + 1) = \dfrac{4x + 1}{x + 1} (x + 1)$ $ x^2 - 4 = 4x + 1$ Subtract $4x + 1$ from both sides: $ x^2 - 4 - (4x + 1) = 4x + 1 - (4x + 1)$ $ x^2 - 4 - 4x - 1 = 0$ $ x^2 - 5 - 4x = 0$ Factor the expression: $ (x - 5)(x + 1) = 0$ Therefore $x = 5$ or $x = -1$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.